Php Echo Mysql Error
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Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up how to display mysql_query error MySql error in php up vote 14 down vote favorite In PHP, I am trying to execute a long MySql query that depends on the user input. However, my query fails with the following message, "Query Failed". Actually I have printed this message whenever the query fails but I am having hard time looking for the reason behind this failure. Unfortunately, I couldn't find it die(mysql_error()) because the error is not specified on the web page. My question is, is there any way to display the error message that caused the failure on the web page. Thank you. Here's my code, $from= "Findings"; $where=""; if($service!= null) { $from = $from . ", ServiceType_Lookup"; $where= "Findings.ServiceType_ID= ServiceType_Lookup.ServiceType_ID AND ServiceType_Name= ". $service; if($keyword!= null) $where= $where . " AND "; } if( $keyword != null) { $where= $where . "Finding_ID LIKE '%$keyword%' OR ServiceType_ID LIKE '%$keyword%' OR Title LIKE '%$keyword%' OR RootCause_ID LIKE '%$keyword%' OR RiskRating_ID LIKE '%$keyword%' OR Impact_ID LIKE '%$keyword%' OR Efforts_ID LIKE '%$keyword%' OR Likelihood_ID LIKE '%$keyword%' OR Finding LIKE '%$keyword%' OR Implication LIKE '%$keyword%' OR Recommendation LIKE '%$keyword%' OR Report_ID LIKE '%$keyword%'"; } $query = "SELECT Finding_ID, ServiceType_ID, Title, RootCause_ID, RiskRating_ID, Impact_ID, Efforts_ID, Likelihood_ID, Finding, Implication, Recommendation, Report_ID FROM ".$from . " WHERE " . $where; echo "wala 2eshiq"; $this->result = $this->db_link->query($query); if (!$this->result) { printf("Query failed: %s\n", mysqli_connect_error()); exit; } $r = mysqli_query($this->db_link, $query); if($r==false) printf("error: %s\n", mysqli_errno($this->db_link)); php mysql share|improve this question asked Sep 1 '12 at 12:16 Traveling Salesman 56131640 You can just use: $this->db_link->error to get the last error mes
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WorkSocial MediaSoftwareProgrammingWeb Design & DevelopmentBusinessCareersComputers Online Courses B2B Solutions Shop for Books San Francisco, CA Brr, it´s cold outside Search Submit RELATED ARTICLES How to Handle MySQL Errors Database Development For Dummies Electronic Health Records For Dummies Microsoft SQL Server 2008 For Dummies Crystal Reports 10 For http://www.dummies.com/programming/databases/how-to-handle-mysql-errors/ Dummies Load more ProgrammingDatabasesHow to Handle MySQL Errors How to Handle MySQL Errors Related Book PHP, MySQL, JavaScript & HTML5 All-in-One For Dummies By Steve Suehring, Janet Valade You use the mysqli functions of the http://www.php5-tutorial.com/mysql/handling-errors/ PHP language, such as mysqli_connect and mysqli_query, to interact with the MySQL database. Things will sometimes go wrong when you use the statements. You may make an error in your typing, such as mistyping a mysql error database name. Sometimes, problems arise that you can't avoid, such as the database or the network being down. You need to include code in your script that handles error situations. You usually want to make your error handling more descriptive to assist with troubleshooting problems during development, but you don't want the extra information displayed to the public. For instance, suppose that you're using an account called root to access php echo mysql your database and you make a typo, as in the following statements: $host = "localhost"; $user = "rot"; $password = "; $cxn = mysqli_connect($host,$user,$password) Because you type "rot" rather than "root", you see a warning message similar to this one: Warning: Access denied for user: 'rot@localhost' (Using password: NO) ... The preceding error message contains the information that you need to figure out the problem -- it shows your account name that includes the typo. However, after your script is running and customers are using it, you don't want your users to see a technical error message that shows your user ID. You want to turn the PHP errors off or send them to an error log file. You could then use a die statement to stop the script and display a polite message to the user, as follows: $cxn = mysqli_connect($host,$user,$password) or die("The Catalog is not available at the moment. Please try again later."); When a mysqli_query() function fails, MySQL returns an error message that contains information about the cause of the failure. However, this message isn't displayed unless you specifically display it. Again, you may want to see these messages when you're developing the script, but you may not want to display them to the public. You
name or a keyword or something like that. By default, PHP will not show you exactly what the problem is, only that you wrote a query which is not entirely correct. Let's try writing a faulty query to see the response from PHP: $query = mysql_query("SELECT id, namme FROM test_users"); while($row = mysql_fetch_array($query)) echo $row['id'] . " - " . $row["name"] . " is from " . $row["country"] . "
"; This is the example we are using a lot in this part of the tutorial, but in this case, we have misspelled the name column to provoke an error, which we get: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in test.php on line 7 As you can see, the error is not thrown until we try using the resource returned by the mysql_query() function, which we do when we call the mysql_fetch_array() function, in my file located on line 7. The error is very generic and not very helpful. This is on purpose, because knowledge about your database structure makes your website more vulnerable to SQL injection attacks, a problem we will discuss later on. You might be able to spot the error and fix it in a lot of situations, but if not, you can use the mysql_error() function to get a bit more information abut the problem. This function simply returns any error returned from the last executed MySQL function. You should only use this function for finding and fixing problems, and then remove it again once the problem has been fixed. Here's the above example, but where we call the mysql_error() function to get more information: $query = mysql_query("SELECT id, namme FROM test_users"); while($row = mysql_fetch_array($query)) echo $row['id'] . " - " . $row["name"] . " is from " . $row["country"] . "
"; echo mysql_error(); This will give you a far more useful error message: Unknown column 'namme' in 'field list' Try making various errors in the SQL query and see the message that MySQL return